JEE Mains 2014 Solved Paper with detailed Answers

This year’s JEE (Joint Entrance Exam) Mains exam test was held on 6th April 2014. Tens of lakh student’s were attended this exam to get admission in IITs and AIEEE colleges. Students search for fully solved paper, questions and detailed solution is going to end now. This JEE question Paper 1 contains 90 questions from Maths, Physics and Chemistry subject streams.

CareerVendor brings full solution of each & every question of JEE 2014 exam. Have a look at the below questions and their detailed solution with answer.

IIT JEE Main 2014 paper solution with answers:

Here are questions from all three paper categories, i.e. Chemistry, Physics and Mathematics. Have a look at the solution provided with reasons. For a reference we are taking Test Booklet CodeH in account.

[divider]

PART – A: PHYSICS

[divider]

1. The pressure that has to be applied to the ends of a steel wire of length 10 cm to keep its length constant when its temperature is raised by 100°C is :

Ques 3. A bob of mass m attached to an inextensible string of length  is suspended from a vertical support. The bob rotates in a horizontal circle with an angular speed w rad/s about the vertical. About the point of suspension :

(1) angular momentum changes in direction but not in magnitude. (2) angular momentum changes both in direction and magnitude. (3) angular momentum is conserved.

(4) angular momentum changes in magnitude but not in direction.

Sol. 1

 

L changes in direction not in magnitude L

Ä V

Ques 4. The current voltage relation of diode is given by I = (e1000V/T – 1) mA, where the applied voltage V is in volts and the temperature T is in degree Kelvin. If a student makes an error measuring ± 0.01 V while measuring the current of 5 mA at 300 K, what will be the error in the value of current in mA ?

(1) 0.5 mA (2) 0.05 mA (3) 0.2 mA (4) 0.02 mA

Sol. (3)

1000 V 5 = e

T – 1

= 6

…(1)

Þ e

1000 V

T

1000 V

Again, I = eT – 1

T

dI = e 1000 V 1000

dV T

1000

T

1000 V

e T dV

Using (1)

DI = 1000 ´ 6 ´ 0.01 = 60 = 60 = 0.2mA

T T 300

Ques 5. An open glass tube is immersed in mercury in such a way that a length of 8 cm extends above the mercury level. The open end of the tube is then closed and sealed and the tube is raised vertically up by additional 46 cm. What will be length of the air column above mercury in the tube now? (Atmospheric pressure = 76 cm of Hg)

Ques 7. A parallel plate capacitor is made of two circular plates separated by a distance of 5 mm and with a dielectric of dielectric constant 2.2 between them. When the electric field in the dielectric is 3 × 104 V/m, the charge density of the positive plate will be close to :

(1) 3 × 104 C/m2 (2) 6 × 104 C/m2

(3) 6 × 10–7 C/m2 (4) 3 × 10–7 C/m2

Sol. 3

By formula of electric field between the plates of a capacitor E = s

Ke0

Þ s = EKe0

= 3 ´104

´ 2.2 ´ 8.85 ´10

-12

= 6.6 ´ 8.85 ´10-8

= 5.841´10-7

@ 6 ´10-7 C/m2

Ques 8. A student measured the length of a rod and wrote it as 3.50 cm. Which instrument did he use to measure it? (1) A screw gauge having 100 divisions in the circular scale and pitch as 1 mm.

(2) A screw gauge having 50 divisions in the circular scale and pitch as 1 mm. (3) A meter scale.

(4) A vernier calliper where the 10 divisions in vernier scale matches with 9 division in main scale and main scale has 10 divisions in 1 cm.

Sol. 4

Least count of vernier calliper is 1 mm = 0.1 mm = 0.01 cm

Ques 9. Four particles, each of mass M and equidistant from each other, move along a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is :

(1) GM (1+ 2 2 ) R(2) 1GM (1+ 2 2 )

2 R

(3) GM R

(4)

2 2 GM R

Sol. 2

Net force on any one particle M

u

GM2

= +

(2R ) 2

GM2

(R 2 ) 2

cos 45° +

GM 2

(R 2 ) 2

cos 45° u

GM 2 é 1 1 ù

ê

ú

= +

2

R 2 ë 4 û

45°

M M

45°

This force will be equal to centripetal force so

Mu 2

GM2 é1 + 2 2 ù u

= ê ú

M

R R 2 ë 4 û uu = GM é1 + 2 2 ù = 1GM (2 2 +1)

4R ë û 2 R

Ques 10. In a large building, there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of 80 W and 1 heater of 1kW. The voltage of the electric mains is 220 V. The minimum capacity of the main fuse of the building will be :

(1) 12 A (2) 14 A (3) 8 A (4) 10 A

Sol. 1

Item No. Power

40 W bulb 15 600 Watt

100 W bulb 5 500 Watt

80 W fan 5 400 Watt

1000 W heater 1 1000 Watt

Total Wattage = 2500 Watt

So current capacity i = P = 2500 = 125 = 11.36 @ 12 Amp.

V 220 11

Ques 11. A particle moves with simple harmonic motion in a straight line. In first t s, after starting from rest it travels a distance a, and in next t s it travels 2a, in same direction, then :

(1) amplitude of motion is 4a (2) time period of oscillations is 6t

(3) amplitude of motion is 3a (4) time period of oscillations is 8t

Sol. 2

A(1 – cos wt) = a

A( 1 – cos 2wt) = 3a

cos wt = æ1 – a ö

ç A ÷

è ø

cos 2wt = æ1 – 3a ö

ç A ÷

2

2 æ1 – a ö è ø

– 1 = 1 – 3a

ç A ÷ A

è ø

Solving the equation

a = 1

A 2

A = 2a

cos wt = 1

2

T = 6t

Sol.

(1) 3A

(2) 6A

(3) 30 mA1

m0 H = m0 ni

(4) 60 mA

3 ´103 = 100 ´ i Þ i = 3A0.1

Ques 12. The coercivity of a small magnet where the ferromagnet gets demagnetized is 3 ´ 103 Am-1. The current required to be passed in a solenoid of length 10 cm and number of turns 100, so that the magnet gets demagnetized when inside the solenoid, is:

Ques 13. The forward biased diode connection is:

(1) 2V 4V (2) -2V +2V

(3) +2V -2V (4) -3V -3V

Sol. 3

By diagram

Ques 14. During the propagation of electromagnetic waves in a medium:

(1) Electric energy density is equal to the magnetic energy density. (2) Both electric and magnetic energy densities are zero.

Ques 15. In the circuit shown here, the point ‘C’ is kept connected to point

‘A’ till the current flowing through the circuit becomes constant. Afterward, suddenly, point ‘C’ is disconnected from point ‘A’ and connected to point ‘B’ at time t = 0. Ratio of the voltage across resistance and the inductor at t = L/R will be equal to :

(1) -1 (2) 1 – e e

A C R B

L

(3) e

1 – e

(4) 1

Sol. 4

Since resistance and inductor are in parallel, so ratio will be 1.

Ques 16. A mass ‘m’ is supported by a massless string wound around a uniform hollow cylinder of mass m and radius R. If the string does not slip on the cylinder, with what acceleration will the mass fall on release?

R

Sol. 4

For the mass m,

mg – T = ma

for the cylinder,

TR = mR 2 a

R

Þ T = ma

Þ mg = 2ma

Þ a = g/2

Ques 17. One mole of diatomic ideal gas undergoes a cyclic process ABC as shown in figure. The process BC is adiabatic. The temperatures at A, B and C are 400 K, 800 K and 600 K respectively. Choose the correct

P statement: (1) The change in internal energy in the process AB is -350 R. (2) The change in internal energy in the process BC is -500 R. (3) The change in internal energy in whole cyclic process is 250 R. (4) The change in internal energy in the process CA is 700 R.

Sol.2

(1) 2gH = nu2(n -2)

(2) gH = (n -2)u2

Sol.

(3) 2gH = n2u21

Time to reach the maximum height

(4) gH = (n -2)2u2

Ques 18. From a tower of height H, a particle is thrown vertically upwards with a speed u. The time taken by the particle, to hit the ground, is n times that taken by it to reach the highest point of its path. The relation between H, u and n is:

t = u

1 g

If t2 be the time taken to hit the ground

-H = ut– 1 gt 2

2 2 2

But t2 = nt1 (given)

nu 1 n 2 u 2

Þ -H = u – g

g 2 g2

Þ 2gH = nu2(n – 2)

Ques 19. A thin convex lens made from crown glass æ m = 3 ö has focal length f. When it is measured in two different

ç 2 ÷

liquids having refractive indices 4

3

relation between the focal lengths is:

è ø

1

and 5 , it has the focal lengths f3

and f2

respectively. The correct

(1) f2 > f and f1 becomes negative (2) f1 and f2 both become negative

(3) f1 = f2 < f (4) f1 > f and f2 becomes negative

Sol. 4

f m = (m1)–

f æ m

ç m

ö

– 1÷

è m ø

– 1

æ 3 öç ÷

Þ f1 = è 2 ø = 4

f æ 3 / 2 – 1ö

ç 4 / 3 ÷

è ø

Þ f1 = 4f

– 1

æ 3 ö

f 2

ç ÷

2 = è ø = -5

f æ 3 / 2 – 1ö

ç 5 / 3 ÷

è ø

Þ f2 < 0

Ques 20. Three rods of Copper, Brass and Steel are welded together to from a Y –shaped structure. Area of cross – section of each rod = 4 cm2. End of copper rod is maintained at 100°C where as ends of brass and steel are kept at 0°C. Lengths of the copper, brass and steel rods are 46, 13 and 12 cms respectively. The rods are thermally insulated from surroundings except at ends. Thermal conductivities of copper, brass and steel are

Ques 21. A pipe of length 85 cm is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below 1250 Hz. The velocity of sound in air is 340 m/s.

Ques 22. There is a circular tube in a vertical plane. Two liquids which do not mix and of densities d1 and d2 are filled in the tube. Each liquid subtends 900 angle at centre. Radius joining their interface makes an angle a with vertical. ratio d1/d2 is

(1) 1 + tan a

1 – tan a

(3) 1 + sin a

1 – sin a

(B) 1 + sin a d2

1 – cos a a

(D) 1 + cos a

1 – cos a

d1

Sol. 1

PA = PB

P0 + d1gR(cos a – sin a) = P0 + d2gR(cosa + sina)

Þ d1 = cos a + sin a = 1 + tan a

d2 cos a – sin a

1 – tan a

23. A green light is incident from the water to the air – water interface at the critical angle (q). Select the correct statement

(1) The spectrum of visible light whose frequency is more than that of green light will come out to the air medium.

(2) The entire spectrum of visible light will come out of the water at various angles to the normal. (3) The entire spectrum of visible light will come out of the water at an angle of 900 to the normal.

(4) The spectrum of visible light whose frequency is less than that of green light will come out to the air

medium.

Sol. 4

As frequency of visible light increases refractive index increases. With the increase of refractive index

critical angle decreases. So that light having frequency greater than green will get total internal reflection and the light having frequency less than green will pass to air.

24. Hydrogen (1H1), Deuterium (1H2), singly ionised Helium (2He4)+ and doubly ionised lithium (3Li6)++ all have one electron around the nucleus. Consider an electron transition from n = 2 to n = 1. If the wave lengths of emitted radiation are l1, l2, l3 and l4 respectively then approximately which one of the following is correct?

(1) l1 = l2 = 4l3 = 9l4 (2) l1 = 2l2 = 3l3 = 4l4

(3) 4l1 = 2l2 = 2l3 = l4 (4) l1 = 2l2 = 2l3 = l4

Sol. 1

1 = Rz2 æ 1 – 1 ö

l ç 12

22 ÷

è ø

\ l = 4

3Rz2

l1 =

l2 =

l3 =

l4 =

4

3R

4

3R

4

12R

4

27R

Þ l1 = l2 = 4l3 = 9l4

25. The radiation corresponding to 3 ® 2 transition of hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field of 3 ´ 10-4T. If the radius of the largest circular path followed by these electrons is 10.0 mm, the work function of the metal is close to :

(1) 0.8 eV (2) 1.6 eV (3) 1.8 eV (4) 1.1 eV

Sol. 4

mv = qBR

KE(max) =

(mv) 2

2m

= 0.8eV

4 9

ê ú

hn = 13.6 é 1 – 1 ùë û

\ W = hn – KE.(max)

= 13.6 5 – 0.8 = 1.1eV

36

Ques 26. A block of mass m is placed on a surface with a vertical cross section given by y = x3/6. If the coefficient of friction is 0.5, the maximum height above the ground at which the block can be placed without slipping is :

(1) 1 m

3

(3) 1 m

6

(2) 1 m

2

(4) 2 m

3

Sol. 3

mg sin q = mmg cosq

tan q = m

dy = tan q = m = 1 dx 2

2

x = 1 , x = ± 1

2 2

y = 1 m .

6

27. When a rubber-band is stretched by a distance x, it exerts a restoring force of magnitude F = ax + bx2 where a and b are constants. The work done in stretching the unstretched rubber band by L is :

aL2

bL3

+ (2)

1 æ aL2

ç

bL3 ö

+ ÷

2 3 2 è

2 3 ø

(3) aL2 + bL3 (4) 1 ( aL2 + bL3)

2

Sol. 1

F = ax + bx2

dw = Fdx

L

W = ò (ax + bx 2 )dx

0

aL2

W =

bL3

+

2 3

28. On heating water, bubbles being formed at the bottom of the vessel detach and rise. Take the bubbles to be spheres of radius R and making a circular contact of radius r with the bottom of the vessel. If r < < R, and the surface tension of water is T, value of r just before bubbles detach is : (density of water is rw)

(1) R 2

(3) R 2

Sol. None

rw g

T

rw g

3T

(2) R 2

(4) R 2

3rw g

T R

rw g

6T

2r

q

(2pr T)sin q =

4 pR 3r .g

3 w

T ´ r ´ 2pr = 4 pR 3r g

R 3 w

4

r2 =

2R rw g

3 T

r = R2 2rwg

3T

29. Two beams, A and B, of plane polarized light with mutually perpendicular planes of polarization are seen through a polaroid. From the position when the beam A has maximum intensity (and beam B has zero intensity), a rotation of Polaroid through 300 makes the two beams appear equally bright. If the initial intensities of the two beams are IA and IB respectively, then IA/IB equals :

(1) 1 (2) 1/3

(3) 3 (4) 3/2

Sol. 2

IA cos230 = IB cos260

IA = 1

IB 3



30. Assume that an electric field E = 30x 2 ˆi exists in space. Then the potential difference VA – VO, where VO is

the potential at the origin and VA the potential at x = 2 m is : (1) -80 J (2) 80 J

(3) 120 J (4) – 120 J

Sol. None



E = 30x 2 ˆi

dV = – -ò E.dx

vA 2

ò dV = -ò 30x 2 dx

v0 0

VA – V0 = – 80 Volt

[divider] PART – B: MATHEMATICS [divider]

Ques 31. The image of the line

x –1 = y – 3 = z – 4

in the plane 2x – y + z + 3 = 0 is the line

(1)

3 1 -5

x + 3 = y – 5 = z – 2

(2)

x + 3 = y – 5 = z + 2

3 1 -5

-3 -1 5

(3)

x – 3 = y + 5 = z – 2

(4)

x – 3 = y + 5 = z – 2

3 1 -5

Sol. 1

-3 -1 5

Line is parallel to plane

Image of (1, 3, 4) is (- 3, 5, 2).

32. If the coefficients of x3 and x4 in the expansion of (1 + ax + bx2) (1 – 2x)18 in powers of x are both zero, then

(a, b) is equal to

(1) æ16 , 251 ö

(2) æ14 , 251 ö

ç 3 ÷

ç 3 ÷

è ø

(3) æ14 , 272 ö

è ø

(4) æ16 , 272 ö

ç 3 ÷

ç 3 ÷

è ø è ø

Sol. 4

1(1 – 2x)18 + ax(1 – 2x)18 + bx2(1 – 2x)18

Coefficient of x3 : (–2)3 18C3 + a(–2)2 18C2 + b(–2) 18C1 = 0

4 ´ (17 ´16 )

(3 ´ 2 )

– 2a × 17 + b = 0

2

….. (i)

Coefficient of x4 : (–2)4 18C4 + a(–2)3 18C3 + b(–2)2 18C2 = 0

( 4 ´ 20 ) – 2a × 16 + b = 0

3

….. (ii)

From equation (i) and (ii), we get

4 æ 17 ´8 – 20 ö + 2a æ 16 – 17 ö = 0

ç 3 ÷ ç 3 2 ÷

è ø è ø

æ 17 ´ 8 – 60 ö

2a ( -19 )

4 ç ÷ + = 0

è 3 ø 6

a = 4 ´ 76´ 6

3 ´ 2 ´19

Þ a = 16

Þ b = 2 ´16 ´16 – 80 = 272

3 3

(1) (-1, 0) È (0, 1) (3) (-2, -1)

(2) (1, 2)

(4) (-¥, -2) È (2, ¥)

Sol.

1

a2 = 3t2 – 2t

For non-integral solution0 < a2 < 1

a Î (- 1, 0) È (0, 1).

33. If a Î R and the equation -3(x – [x])2 + 2 (x – [x]) + a2 = 0 (where [x] denotes the greatest integer £ x) has no integral solution, then all possible values of a lie in the interval

1

[Note: It is assumed that a real solution of given equation exists.] (0, 0) (2/3,0)

  

      2

ë û ë û

34. If éa ´ b b ´ c c ´ a ù = l éa b c ù , then l is equal to

(1) 2 (2) 3 (3) 0 (4) 1

Sol. 4



        2

ë û ë û

éa ´ b b ´ c c ´ a ù = éa b c ù

l = 1.

35. The variance of first 50 even natural numbers is

(1) 833

4

(2) 833

(3) 437 (4) 437

4

Sol. 2

s2 = ç åxi ÷ – x 2

æ 2 ö

è n ø

50

å 2r

x = r =1 = 51

50

50

å 4r 2

s2 = r =1 – (51)2 = 833

50

36. A bird is sitting on the top of a vertical pole 20 m high and its elevation from a point O on the ground is

45°. It flies off horizontally straight away from the point O. After one second, the elevation of the bird from

O is reduced to 30°. Then the speed (in m/s) of the bird is

(1) 40 (

2 – 1)

(2) 40 ( 3 – 2 )

(3) 20 2 (4) 20 (

Sol. 4

tan 30º = 20 = 1

20 + x 3

20 + x = 20 3

3 – 1)

20m = h

20m

x = 20 (

3 -1)

p/4 p/6

Þ Speed is 20 (

3 -1) m/sec.

0 20m x

p

37. The integral ò

0

1 + 4 sin 2 x – 4 sin x

2 2

dx equals

(1) p – 4 (2) 2p – 4 – 4 3

3

(3) 4 3 – 4

Sol. 4

(4) 4 3 – 4 – p

3

p

I = ò

0

1 + 4 sin 2 x – 4 sin x dx

2 2

y = 2 sin x/2

2

p

1

x

= ò 1 – 2 sin 2 dx

0

p / 3

=

æ1 – 2 sin x ö dx +

p

æ 2 sin x – 1ö dx p p

ò è ø

ò è ø

ç 2 ÷ ç 2 ÷

0 p/3 3

= æ x + 4 cos x ö

p / 3

p

+ æ -4 cos x – x ö

ç ÷ ç ÷

è 2 ø 0 è

2 ø p / 3

= – p + 8 × 3 – 4

3 2

= 4 3 – 4 – p

3

Ques 38. The statement ~(p « ~q) is

(1) equivalent to p « q (2) equivalent to ~p « q

(3) a tautology (4) a fallacy

Sol. 1

P

q

~ q

p « ~ q

~ (p « ~ q)

p « q

T

T

F

F

T

T

T

F

T

T

F

F

F

T

F

T

F

F

F

F

T

F

T

T

39. If A is an 3 ´ 3 non-singular matrix such that AA¢ = A¢A and B = A-1A¢, then BB¢ equals

(1) I + B (3) B-1

(2) I(4) (B-1)¢

Sol.

2B = A-1A¢ Þ AB = A¢

ABB¢ = A¢B¢ = (BA)¢ = (A-1A¢A)¢ = (A-1AA¢)¢ = A.

Þ BB¢ = I.

40. The integral

1

æ1 + x – 1 ö e x+ x dx is equal to

ò è x ø

ç ÷

x + 1

x+ 1

(1) ( x – 1) e

(3) ( x + 1) e

Sol. 2

x + c (2) x e

x + 1

x + c (4) – x e

æ x + 1 ö

x + c

x+ 1

x + c

1 + x – 1 eç

x ÷ dx

æ ö è ø

ò ç x ÷

è ø

æ x + 1 ö

1 æ x + 1 ö

= eè

x ø dx +

x æ1 –

ö eè

x ø dx

ç ÷ ç ÷

ò ò ç ÷

è x 2 ø

x + x + x +

æ 1 ö æ 1 ö æ 1 öç ÷ ç ÷ ç ÷

= ò eè

x ø dx + xeè

x ø – ò eè

x ø dx

x +

æ 1 öç ÷

= xeè

x ø + c

41. If z is a complex number such that |z| ³ 2, then the minimum value of

z + 1

2

(1) is equal to 5

2

(3) is strictly greater than 5

(2) lies in the interval (1, 2)

(4) is strictly greater than 3 but less than 5

Sol. 2

|z| ³ 2

z + 1 ³

z – 1

2 2 2

³ 2 – 1 ³ 3 .

2 2 2 2

Hence, minimum distance between z and æ – 1 , 0 ö is 3

ç 2 ÷

è ø 2

42. If g is the inverse of a function f and f ¢(x) = 1 , then g¢(x) is equal to

69. The octahedral complex of a metal ion M3+ with four monodentate ligands L1, L2, L3 and L4 absorb wavelengths in the region of red, green, yellow and blue, respectively. The increasing order of ligand strength of the four ligands is:
absorption of higher energy radiation.

¾¾¾¾¾¾¾¾

V I B G Y O R ® decrea sin g energy which is accompanied by the

70. For the estimation of nitrogen, 1.4 g of organic compound was digested by Kjeldahl method and the